## 10-Digit Numbers

1. Find a 10 digit number such that:
• the first digit tells how many zeros are in the number,
• the second digit tells how many ones are in the number,
• the third digit tells how many twos are in the number, etc.
2. Find the smallest 10 digit number that uses all the digits 0-9 and is divisible by 1 through 10.
3. Find the smallest 10 digit number that uses all the digits 0-9 and:
• the first digit is divisible by 1,
• the first 2 digits (taken as a 2-digit number) are divisible by two,
• the first 3 digits (taken as a 3-digit number) are divisible by three, etc.

#### Source: 1,3. Martin Gardner. 2. Pi Mu Epsilon Math Journal. Taken from "Ken's Puzzle of the Day", May 5, 1994

Solution:
1) 6210001000
Solvers:
Becky Fabie-Albert (becky10g@webquest.com),
Suresh Gopalakrishnan (sureshgk@inf.com),
Rick Simineo (Rick@plains.uwyo.edu)
2) 1234759680
Solvers: Becky Fabie-Albert, Rick Simineo
3) 3816547290
Solvers: Becky Fabie-Albert

(1) is rather straight-forward to find: There must be mostly 0's in the number, so the count of 0's is high, and there is then a 1 somewhere later in the number. Since there is a 1 later, the value for the count of 1's in non-zero, and must be greater than 1. Trying 2, with a 1 in the 2's position leads to the solution above.
(2) can be found by brute force to be the smallest. (I used a computer.)
(3) takes a little more work. The last digit must be zero. The fifth digit must be 5. The odd digits are in the odd positions. So, the fourth digit must be 2 or 6, and the eighth digit is 6 or 2. The second trio of digits must be divisible by three. The possible trios are 254, 258, 654, 658. Only 258 and 654 are divisible by three. At this point, the two options for the number are:
.4.258.6.0
.8.654.2.0
Digits six thru eight are divisible by 8, so the new options become:
.4.25816.0, .4.25896.0
.8.65432.0, .8.65472.0
The first three digits are divisible by three. For the first option above, there is no combination of 3,7 and 9 with 4 that will give a number divisible by 3. For the rest, there are the following possibilities: 1472589630, 7412589630
1896543270, 9816543270
1836547290, 3816547290
Now, all n-length numbers have been satisfied except for n=7. We need merely check for divisibility by 7 on each of the first seven digits of the numbers above. Doing this finds only one of the answers works: 3816547290

Becky Fabie-Albert's solution for (3):
To derive the 3rd 10-digit number, I used the Rules for Divisibility :
A number is divisible -
by 2 if it ends in 0,2,4,6 or 8;
by 3 if the sum of its digits is divisible by 3;
by 4 if the number formed by its last two digits is divisible by 4;
by 5 if it ends in 0 or 5;
by 6 if it is even and the sum of its digits is divisible by 3;
by 8 if the number formed by its last 3 digits is divisible by 8;
by 9 if the sum of its digits is divisible by 9;
by 10 if it ends with 0.

Given these rules, we can establish that the 10th digit is '0'.
So the 5th digit must be '5'.  We also know that the 2nd, 4th, 6th and
8th digits must be even, and therefore the 1st, 3rd, 7th and 9th must
be odd.

Using the Divisible-by-4 Rule, we find that the 4th digit must be
either '2' or '6', since the 3rd digit is '1', '3', '7' or '9'.
(There is no 2-digit number with 1st digit '1', '3', '7' or '9'
and 2nd digit '4' or '8' that is divisible by 4).

Similarly, the 8th digit must also be either '2' or '6'. (A number that
is divisible by 8 must also be divisible by 4, so the Divisible-by-4
Rule applies to the 8-digit number).

Given that the 4th and 8th digits are '2' and '6' (in some order),
the 2nd and 6th digits must be '4' and '8' (in some order).

Since we are looking for the SMALLEST 10-digit number, first assume the
2nd digit is '4' and therefore the 6th digit is '8'.
Using the Divisible-by-3 Rule, we find that the 1st and 3rd digits must
be, in some order, '1' and '7'.
Using the Divisible-by-6 Rule, we find that the 4th digit must be '2'.
So our valid combinations are:
A) 1 4 7 2 5 8 3 6 9 0
B) 1 4 7 2 5 8 9 6 3 0
C) 7 4 1 2 5 8 3 6 9 0
D) 7 4 1 2 5 8 9 6 3 0
Checking for divisibility by 7 and 8, we find that none of the above
combinations are valid. (Combination A is not divisible by 8 and
Combinations B,C,D are not divisible by 7).

Therefore are assumption that the 2nd digit = '4' is false.
So the 2nd digit must be '8' and the 6th digit must be '4'.
And the 4th and 8th digits, in some order, must be '2' and '6'.

If we assume that 4th digit is '2', we find that there are no possible
combinations that satisfy both the Divisible-by-3 and Divisible-by-6
Rules.
Therefore the 4th digit is '6' and the 8th digit is '2'.
So the number is :   ? 8 ? 6 5 4 ? 2 ? 0.

To satisfy the Divisible-by-8 Rule, we find the 7th digit must be
either '3' or '7'.
So the possible valid combinations are:
A) 1 8 7 6 5 4 3 2 9 0
B) 1 8 9 6 5 4 3 2 7 0
C) 7 8 1 6 5 4 3 2 9 0
D) 7 8 9 6 5 4 3 2 1 0
E) 9 8 1 6 5 4 3 2 7 0
F) 9 8 7 6 5 4 3 2 1 0
G) 1 8 3 6 5 4 7 2 9 0
H) 1 8 9 6 5 4 7 2 3 0
I) 3 8 1 6 5 4 7 2 9 0
J) 3 8 9 6 5 4 7 2 1 0
K) 9 8 1 6 5 4 7 2 3 0
L) 9 8 3 6 5 4 7 2 1 0
Checking for divisibility by 3, we can rule out combinations A,C,J,L.
Checking for divisibility by 7, we can rule out combinations B,D,E,F,G,H,K.

So "3 8 1 6 5 4 7 2 9 0" is the ONLY answer.

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